【CodeForces】35E Parade 线段树

E. Parade

time limit per test

4 seconds

memory limit per test

64 megabytes

input

input.txt

output

output.txt

No Great Victory anniversary in Berland has ever passed without the war parade. This year is not an exception. That’s why the preparations are on in full strength. Tanks are building a line, artillery mounts are ready to fire, soldiers are marching on the main square... And the air forces general Mr. Generalov is in trouble again. This year a lot of sky-scrapers have been built which makes it difficult for the airplanes to fly above the city. It was decided that the planes should fly strictly from south to north. Moreover, there must be no sky scraper on a plane’s route, otherwise the anniversary will become a tragedy. The Ministry of Building gave the data onn sky scrapers (the rest of the buildings are rather small and will not be a problem to the planes). When looking at the city from south to north as a geometrical plane, thei-th building is a rectangle of height h_i. Its westernmost point has the x-coordinate ofl_i and the easternmost — ofr_i. The terrain of the area is plain so that all the buildings stand on one level. Your task as the Ministry of Defence’s head programmer is to find anenveloping polyline using the data on the sky-scrapers. The polyline’s properties are as follows:

• If you look at the city from south to north as a plane, then any part of any building will be inside or on the boarder of the area that the polyline encloses together with the land surface.
• The polyline starts and ends on the land level, i.e. at the height equal to 0.
• The segments of the polyline are parallel to the coordinate axes, i.e. they can only be vertical or horizontal.
• The polyline’s vertices should have integer coordinates.
• If you look at the city from south to north the polyline (together with the land surface) must enclose the minimum possible area.
• The polyline must have the smallest length among all the polylines, enclosing the minimum possible area with the land.
• The consecutive segments of the polyline must be perpendicular.

Picture to the second sample test (the enveloping polyline is marked on the right).

Input

The first input line contains integer n (1 ≤ n ≤ 100000). Then follown lines, each containing three integers h_i, l_i, r_i (1 ≤ h_i ≤ 10⁹,  - 10⁹ ≤ l_i < r_i ≤ 10⁹).

Output

In the first line output integer m — amount of vertices of the enveloping polyline. The next m lines should contain 2 integers each — the position and the height of the polyline’s vertex. Output the coordinates of each vertex in the order of traversing the polyline from west to east. Remember that the first and the last vertices of the polyline should have the height of 0.

Sample test(s)

Input

23 0 24 1 3

Output

60 00 31 31 43 43 0

Input

53 -3 02 -1 14 2 42 3 73 6 8

Output

14-3 0-3 30 30 21 21 02 02 44 44 26 26 38 38 0

传送门：【CodeForces】35E Parade

题目大意：在二维平面上，给你每个建筑的左右坐标以及高度，问你建筑外部轮廓的转折点。

题目分析：离散化+线段树随便搞。

PS：昨天就写好了，然后一直错在第一组数据简直不明觉厉，后来发现早期CF竟然是文件输入输出！然后我不知道还以为CF坏了，顿时感觉自己萌萌哒QUQ。。。

代码如下：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std ;#define ls ( o << 1 )#define rs ( o << 1 | 1 )#define rt o , l , r#define root 1 , 1 , n#define lson ls , l , m#define rson rs , m + 1 , r#define mid ( ( l + r ) >> 1 )#define clear( A , X ) memset ( A , X , sizeof A )const int maxN = 300005 ;struct Line {int l , r , h ;Line () {}void input () {scanf ( "%d%d%d" , &h , &l , &r ) ;}bool operator < ( const Line &t ) const {return h < t.h ;}} ;struct Node {int x , y ;} ;Line line[maxN << 1] ;Line seg[maxN << 1] ;Node ans[maxN << 1] ;int set[maxN << 2] ;int a[maxN << 2] ;int Unique ( int a[] , int n ) {int cnt = 1 ;sort ( a + 1 , a + n + 1 ) ;for ( int i = 2 ; i <= n ; ++ i ) if ( a[i] != a[cnt] ) a[++ cnt] = a[i] ;return cnt ;}int Again_Unique ( Line line[] , int n ) {int cnt = 1 ;for ( int i = 2 ; i <= n ; ++ i ) {if ( line[i].h == line[cnt].h ) {line[cnt].r = line[i].r ;}else line[++ cnt] = line[i] ;}return cnt ;}int Search ( int x ,  int l , int r ) {while ( l < r ) {int m = mid ;if ( a[m] >= x ) r = m ;else l = m + 1 ;}return l ;}void PushDown ( int o ) {if ( set[o] ) {set[ls] = set[rs] = set[o] ;set[o] = 0 ;}}void Update ( int L , int R , int v , int o , int l , int r ) {if ( L <= l && r <= R ) {set[o] = v ;return ;}PushDown ( o ) ;int m = mid ;if ( L <= m ) Update ( L , R , v , lson ) ;if ( m <  R ) Update ( L , R , v , rson ) ;}void Query ( int o , int l , int r ) {if ( l == r ) {line[l].l = a[l] ;line[l].r = a[r + 1] ;line[l].h = set[o] ;return ;}PushDown ( o ) ;int m = mid ;Query ( lson ) ;Query ( rson ) ;}void work () {//题目需要文件输入输出freopen ( "input.txt" , "r" , stdin ) ;freopen ( "output.txt" , "w" , stdout ) ;int n , cnt = 0 ;while ( ~scanf ( "%d" , &n ) ) {clear ( set , 0 ) ;for ( int i = 1 ; i <= n ; ++ i ) {seg[i].input () ;a[++ cnt] = seg[i].l ;a[++ cnt] = seg[i].r ;}cnt = Unique ( a , cnt ) ;sort ( seg + 1 , seg + n + 1 ) ;for ( int i = 1 ; i <= n ; ++ i ) {int L = Search ( seg[i].l , 1 , cnt + 1 ) ;int R = Search ( seg[i].r , 1 , cnt + 1 ) ;Update ( L , R - 1 , seg[i].h , 1 , 1 , cnt - 1 ) ;}Query ( 1 , 1 , cnt - 1 ) ;cnt = Again_Unique ( line , cnt - 1 ) ;int __cnt = 0 ;ans[__cnt].x = line[1].l ;ans[__cnt].y = 0 ;++ __cnt ;for ( int i = 1 ; i <= cnt ; ++ i ) {ans[__cnt].x = line[i].l ;ans[__cnt].y = line[i].h ;++ __cnt ;ans[__cnt].x = line[i].r ;ans[__cnt].y = line[i].h ;++ __cnt ;}ans[__cnt].x = line[cnt].r ;ans[__cnt].y = 0 ;printf ( "%d\n" , __cnt + 1 ) ;for ( int i = 0 ; i <= __cnt ; ++ i ) printf ( "%d %d\n" , ans[i].x , ans[i].y ) ;}}int main () {work () ;return 0 ;}